Existence of a finite group with a given decomposition for a tensor square of one irreducible complex representation

Question

In this post, irrep and dim mean "irreducible complex representation" and "dimension", respectively. It would be helpful (in a problem of monoidal category) to find a finite group $G$ with (at least) two irreps of dim $5$ (denoted $5_1$ and $5_2$) and (at least) two irreps of dim $7$ (denoted $7_1$ and $7_2$) with $$ 5_1 \otimes 5_1 \simeq 1 \oplus 5_1 \oplus 5_2 \oplus 7_1 \oplus 7_2$$Question: Is there such a finite group $G$?

Remark: such group should be of order a multiple of $35$ and should admit irreps of dims $5$ and $7$, which is helpful to rule out the following cases with GAP on a laptop:

• simple, of order less than $10^6$,
• perfect, of order less than $15120$,
• general, of order less than $2240$.

Let $a_n$ be the smallest order of a group with an irrep of dim $n$ (oeis.org/A220470): $1, 6, 12, 20, 55, 42, 56, 72, 144, 110, 253, 156, 351, 336, 240, 272,\dots$

In particular, $a_5=55$ (given by $C_{11} : C_5$) and $a_7=56$ (given by $C_2^3 : C_7$). Now, I don't even know if there exists a group $G$ with $|G|<55 \times 56=3080$, and with irreps of dims $5$ and $7$.

Answer 1

I think that there is indeed no such finite group $G$, whether simple or otherwise. Note first that the representation $5_{1}$ can be assumed to be faithful ( for if $K$ is its kernel, then the group $G/K$ has the same property), so from now on, we assume it faithful.

Note next that $Z(G) = 1$, since if $5_{1}$ lies over a linear character $\lambda$ of $Z(G)$, then we must have $\lambda^{2} =1 $ since the trivial character occurs in $5_{1} \otimes 5_{1}$. However $\lambda = \lambda^{2}$ since $5_{1}$ also occcurs in $5_{1} \otimes 5_{1}$. Hence $\lambda$ is trivial.

Now $ N = O_{5^{\prime}}(G)$ is Abelian by Clifford's Theorem. If $N$ is non-trivial, then it is also non-central, sincee $Z(G) = 1$, and it follwws from Clifford's Theorem that the representation $5_{1}$ is (up to equivalence) monomial. Then $G$ has an Abelian normal subgroup $A$ such that $G/A$ is isomorphic to a subgroup of $S_{5}$. But in that case, $G$ has an Abelian normal Sylow $7$-subgroup, and $G$ has no irreducible character of degree $7$ (by a theorem of Ito, the degree of a complex irreducible character of $G$ divides $[G:A]$ whenever $A \lhd G$ is Abelian). Hence it follows that $N = 1$. More generally, this argument shows that the representation $5_{1}$ is primitive, ie not (equivalent to one) induced from any proper subgroup of $G$.

There are several ways to finish from here. One is to invoke Brauer's classification of the finite primitive subgroups of ${\rm GL}(5, \mathbb{C})$ and note that none of these has tthe order of $G/Z(G)$ divisible by $7$.

Another is to note that if $O_{5}(G)$ is non-trivial, then it is irreducibly represented by $5_{1}$, in which case $Z(G)$ has order divisible by $5$, a contradiction.

Now we are reduced to the case $F(G) = 1$ and we continue until we see that $M = F^{\ast}(G)$ is a finite simple subgroup of ${\rm GL}(5,\mathbb{C})$ of orer divisible by $35$. But by a theorem of Feit, if $G$ is a finite simple subgroup (of order divisible by the prime $p$) of ${\rm GL}(p-2,\mathbb{C})$ for some prime $p$, then $p$ is a Fermat prime and $G = {\rm SL}(2,p-1)$ ( we may apply this with $p = 7$, so obtain a contradiction).